Recent content by gungo

I learned in class that 0.25 m was the near point of a normal eye

I found the power of the eye itself by doing P= 1/.25 m + 1/ -.75 m. I used negative .75 m because the image would be on the same side of the object. I got a power of 2.67 d. The I added that to the .75 d of the lens. Then I used the same equation and did 3.42= 1/.25 m +1/q to find q, and got...

Homework Statement A dose of 5 Gy of X-rays is required to destroy 63% of a particular bacterial population. It requires 1 Gy of beta particles of mean energy 1.0 MeV to accomplish the same result. What is the wR for the beta particles in this population of bacteria? Homework Equations H= (wR)...

thank you!

Homework Statement The count-rate from a gamma source is measured to be 1000 counts per minute. When 1.0 cm of lead is placed between the source and the detector, the count rate is reduced to 100 counts per minute. What additional thickness of lead would have reduced the count-rate to 50 counts...

It should be on the opposite side, so q is negative,right? That would give the right answer.

Homework Statement A hyperopic eye has a near point of 1.10 m. Calculate the focal length of the spectacle lens required to correct the hyperopia (assume that the near point of the normal eye is 25.0 cm.) Homework Equations 1/f=1/p+1/q The Attempt at a Solution I just used .25 m as p and...

Whoops I forgot about using algebra. I got it, thanks!

Homework Statement Two resistors are connected in parallel. The total resistance of the combination is 6.0 ohms. If one of the resistors is 10 ohms, what is the value of the other? Homework Equations 1/RT= 1/R1+1/R2 The Attempt at a Solution I got 15 just by doing trial and error, and I'm...

Homework Statement Students used a speed gun to measure the final speed of slider at the bottom of the slide. They measure the height difference between the top and bottom of the slide to be 3.0 m. If the speed of a slider at the bottom of the slide is 5.0 m/s, what is the efficiency of the...

Homework Statement I know that on a horizontal surface. if the net force is 0 in the y direction the weight of object=the normal force, but wouldn't the normal force always equal the magnitude of gravity on an object, just in opposite directions? Because if the weight is stronger than the...

whoops sorry

dy:0 dx:250 v2^2=v1^2+2ad ?=?+2(-9.8)(0) t:? t:? v2=v1+at ?=?+-9.8? a:-9.8 a:0 d=v1(t)+1/2(a)(t)^2 0=?(?)+1/2(-9.8)(?)^2 vy1...

So dy isn't 250 tan 17...it's 0 because it lands on the ground. But then we have nothing to input into the equations?

Sorry I don't understand